I am not quite sure how to show that $5^{2m} ≡ 2^{2m} \bmod 7$ , but not $5^{2m} ≡ -2^{2m} \bmod 7$
My answers always show that $5^{2m} ≡ -2^{2m} \bmod 7$ , however, when I denote m with an actual number, it never works.
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1It would be helpful, if you post your answer. My guess is that you mixed $(-2)^{2m}$ with $-2^{2m}$ – Peter Sep 21 '20 at 07:58
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if this is true, how is 5^2m+1 ≡ -2^2m+1, if its not (-2) – Sep 21 '20 at 08:02
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We have $$5^{2m+1}\equiv (-2)^{2m+1}=-2^{2m+1}$$ because of the odd exponent. – Peter Sep 21 '20 at 08:04
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just to clarify, is -2^2m+1 = -(2^2m+1)? – Sep 21 '20 at 08:08
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Yes , it is. Without parantheses , the minus-sign applies after the power has been evaluated. – Peter Sep 21 '20 at 08:09
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1Please don't rollback improvements to your question, esp. those that remove notational ambiguities. – Bill Dubuque Sep 21 '20 at 08:25
2 Answers
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$$5^2\equiv2^2\pmod7$$
$$\implies(5^2)^m\equiv(2^2)^m$$ for any integer $m\ge0$
Now if $5^{2m}\equiv-2^{2m}\pmod7$
We need $2^{2m}\equiv-2^{2m}$
As $(2,7)=1,$ this implies $$1\equiv-1\pmod7\iff2\equiv0$$
lab bhattacharjee
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$\bmod 7\!:\,\ \color{#c00}{5\equiv -2}\Rightarrow\, \color{#c00}5^{2m}\!\equiv (\color{#c00}{-2})^{2m}\!\equiv 2^{2m} := n,\,$ and $\,n\not\equiv -n,\,$ else $\,2n\equiv 0\overset{\large \times\ 4}\Longrightarrow n\equiv 0$
Bill Dubuque
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We applied the Congruence Power Rule in the first arrow. Also note $,n\equiv 0\Rightarrow 7\mid 2^{2m}\Rightarrow\ 7\mid 2,$ by Euclid's Lemma. – Bill Dubuque Sep 21 '20 at 08:05